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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 1

Lesson 13: Limits from equations (factoring & rationalizing)# Finding limits by factoring (cubic)

Sal finds the limit of (x³-1)/(x²-1) at x=1 by factoring and simplifying the expression. Created by Sal Khan.

## Want to join the conversation?

- Where can I watch a tutorial on the kind of division Sal does at1:40?(104 votes)
- is x^3 - y^3 = (x-y)(x² + xy + y²) ?(15 votes)
- Yes, if you need proof:

(x-y)^3=x^3 - y^3 - 3xy(x-y)

=> (x - y)^3 + 3xy(x-y) = x^3 - y^3

=> x^3 - y^3 = (x - y) [(x - y)^2 + 3xy]

=> x^3 - y^3 = (x - y)(x^2 + y^2 - 2xy + 3xy)

=> x^3 - y^3 = (x - y)(x^2 + xy + y^2 )(24 votes)

- At3:13, Sal writes when x =/= 1. But after simplification, even x=-1 will give an undefined answer. Why do we not take that into consideration? How do we know what to put conditions for and what not to put conditions for.

I understand that since the question deals with limit of x going to 1, -1 is not that important, but my question is still how do we generally know what to put conditions for.(10 votes)- He only stated the restriction for the factor that cancelled. Since it cancelled it is no longer obvious so you must state it explicitly.

Since the x+1 factor is still there, the x=/= -1 restriction is obvious by looking at the problem.(10 votes)

- how do I do the problem:

Find the limit:

lim as x approaches 0: {1/(3+x)}-(1/3)/x

fractions in the numerator...I forget this algebra/simplifying(5 votes)- The problem you have is that you don't want to divide by 0, right? So lets try to get rid of the x'es in the numerator, because if we do that we wont have any problem with letting x go to 0 am I right?

So my suggestion would be multiplying both denominator and numerator with x and see what happens! :)(7 votes)

- I read a popular anomaly in maths where the end result is 2=1 when we take A=B.

It goes like this- Let A=B

=> A^2 =AB => A^2 - B^2 = AB - B^2 => (A+B)(A-B) = B (A-B) ------ Step n

=> A+B=B

=> 2B=B => 2=1

The explanation given there was that in step n we divided both sides by zero. As A-B=0, therefore this anomaly came up.

Here too aren't we doing the same thing?(5 votes)- Good question. At3:00, Sal is careful to point out that the division he is performing is valid only if we assume x does not equal 1. We make that assumption when performing a limit calculation because we aren't trying to find the value when x = 1. Instead, we're trying to find the value that is approached as x gets closer and closer to 1. As long as we allow for some difference between x and 1, we aren't dividing by zero and the operation is valid.

Now it may seem like a contradiction that Sal goes on after that to compute the limit by plugging 1 into the resulting fraction, but again you have to keep in mind that he isn't calculating the value of the fraction at 1, he's calculating the value of the limit at 1, which means finding the value that the surrounding points gravitate towards. Plugging in 1 at this point gives us that value.(7 votes)

- Why was he allowed to use 1 if he said x cannot equal one?(4 votes)
- He was allowed to substitute 1 for x even though there was a restriction of "x does not equal 1" because he was finding the
*limit*of the original expression as x*approaches*1. To find the limit, he can substitute 1 for x into the new expression for which x=1 is defined to get the limit for the old expression for which x=1 is not defined. However, this is*not*the value of the old expression at x=1: It is the limit. That's why he can substitute 1 for x into the new expression despite the restriction.(8 votes)

- Would anyone able to show me why the following limit is 9? I would be much appreciate if steps are also shown. Thanks!!

Lim (sin ^2 (3x)) / x^2

x -> 0(3 votes)- (In what follows, it is assumed that
`x → 0`

in every limit statement, if not otherwise stated.)

One way to prove the identity is by making use of the following preliminary relations:`1. lim ( sin(x) / x ) = 1;`

`2.`

If`lim ƒ(x) = F`

and`lim g(x) = G`

, both as`x → a`

, then`lim ƒ(x)g(x) = FG`

as`x → a`

, where`a`

is any real number.

The fact that`lim ( sin²(3x) / x² ) = 9`

may now be deduced by rewriting`sin²(3x) / x²`

to a form we recognise.`I)`

Properties`1.`

and`2.`

imply that`lim ( sin²(x) /`

`x² ) = 1`

. To see this, observe that`lim sin²(x) / x² = lim ( sin(x) / x )² = lim ( sin(x) / x ) · ( sin(x) / x )`

. By properties`1.`

and`2.`

, we have, then,`lim sin²(x) / x² = (lim sin(x) / x) · (lim sin(x) / x) = 1 · 1 = 1`

.`II)`

We now compute`lim sin²(3x) / (3x)²`

. Let`u = 3x`

. Then`u → 0`

as`x → 0`

; hence`lim sin²(3x) / (3x)² = lim sin²(u) / u² = 1`

, which follows from`I)`

.`III)`

Observe that`sin²(3x) / x² = sin²(3x) / ([1/9] · 9x²) = 9 · ( sin²(3x) / (3x)² )`

. By properties`II)`

and`2.`

, it follows that`lim sin²(3x) / x²`

exists and equals`(lim 9) · (lim sin²(3x) / (3x)²) = 9 · 1 = 9`

.

This completes the proof.(5 votes)

- If we differentiate the numerator and denominator by L'hospital rule that would give the same answer right?(3 votes)
- If you are referring to the problem in the video, that is a 0/0 form and thus is valid for l'Hopital's. So, yes, you could do it that way (and I would definitely solve it that way)

lim x→1 (x³-1)/(x²-1)

= lim x→1 (3x²)/(2x)

= ³⁄₂

However, you should not discount the techniques Sal used in the video because you cannot use l'Hopital's on many difficult limits, so it is good to know the techniques for finding the limits without l'Hopital's.

Furthermore, even when you can use l'Hopital's, it is not always the easiest method.(6 votes)

- at0:58how does x^2-1 becomes (x-1)(x+1)?(1 vote)
- I'm sorry, but if you have not yet had enough algebra to know how to factor a difference of squares, then you need to go back and study that material before proceeding to study calculus.

Studying calculus requires that you have already mastered at the very least Algebra I and II and basic Geometry. Without mastering those, you will not be able to understand the content in a Calculus course. Preferably, you would also have had a Trigonometry / Pre-calculus course.

Calculus requires mastery of the ability to factor polynomials and rational expressions, so it will not explain how to do this. Kahn Academy has videos on factoring quadratic expressions that you should review.

https://www.khanacademy.org/math/algebra2/polynomial_and_rational/quad_factoring/v/factoring-quadratic-expressions(5 votes)

- Isn't (x-1)/(x-1) = 0/0 ? Are you not canceling 0 with 0?(1 vote)
- As long as x≠1, (x-1)/(x-1) = 1

EG: x=5

(x-1)/(x-1) = 4/4 = 1

Try with any number, answer is 1.

BUT IF x = 1;

(x-1)/(x-1) = 0/0 = undefined, indeterminate, etc.(4 votes)

## Video transcript

Let's try to find the
limit as x approaches 1 of x to the third minus
1 over x squared minus 1. And at first when you just
try to substitute x equals 1, you get 0/0 1 minus
1 over 1 minus 1. So that doesn't help us. So let's see if we can try
to simplify this in some way. So you might
immediately recognize-- so let's rewrite this
expression right over here so it's x to the third
minus 1 over x squared minus 1. This on the bottom
immediately jumps out as a difference of squares. So we know on the
bottom that this could be factored as x
minus 1 times x plus 1. And so if somehow
this thing on the top also has an x minus 1 as a
factor, then that x minus 1 will cancel with
this, and then we're not going to have an
issue of dividing by 0. The reason why I care
about the x minus 1 term is that this is what's making
our denominator equal 0. When you say x equals 1, you
have 1 minus 1 times 1 plus 1. So 0 times 2, it's this 0
that's making our denominator 0. So if we can have an
x minus 1 up here, then we can cancel these out
for any x not equal to 1. And then we might have
a much simpler thing to find the limit of. So let's think about whether
x to the third minus 1 is the product of x minus
1 and something else. And to do that we
can do a little bit of algebraic long division. Some of you guys might already
recognize a pattern here, but we'll try to do-- well,
let's divide x minus 1 into it to see whether
it divides evenly into x to the third minus 1. So x minus 1-- we just
look at the highest degree term-- x goes into x to
the third x squared times. Goes x squared times. Actually, let me
do it this way so that way we can keep
track of the place. So this would be x-- this
would be the second degree place, first degree place, and
this would be the constant. So x to the third minus 1. x goes into x to the
third x squared times. x squared times x
is x to the third. x squared times negative
1 is minus x squared. And now we're going to
want to subtract this. So we are then left
with x squared. x goes into x squared
x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're
going to subtract this. We'll swap the signs,
negative and positive. And so these cancel out,
and we're left with x. And then we bring
down a minus 1. x minus 1 goes into x
minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and
then you have no remainder. So this numerator
right over here can be factored as x minus 1
times x squared plus x plus 1. And so we can say that this
is the same exact thing. We can have these cancel out if
we assume x does not equal 1. So that is equal to x squared
plus x plus 1 over x plus 1, for x does not equal 1. And that's completely
fine, because we're not evaluating x equals 1. We're evaluating
as x approaches 1. So this is going to be the
same thing as the limit as x approaches 1 of x squared
plus x plus 1 over x plus 1. And now this is
much easier to find. You could literally
just say, well, what happens as we get
right to x equals 1? Then you have 1 squared,
which is 1 plus 1 plus 1, which is 3, over
1 plus 1, which is 2. So we get that equaling 3/2.